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Q.
The value of the following determinant $ \left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right| $ is:
Bihar CECEBihar CECE 2004
Solution:
Let $ \Delta =\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \\ \end{matrix} \right| $
Applying $C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1}$
$ =\left| \begin{matrix} 1 & 0 & 0 \\ {{a}^{3}} & b-a & c-a \\ {{a}^{3}} & {{b}^{3}}-{{a}^{3}} & {{c}^{3}}-{{a}^{3}} \\ \end{matrix} \right| $
$ =(b-a)(c-a) $
$ \times \,\,\left| \begin{matrix} 1 & 0 & 0 \\ a & 1 & 1 \\ {{a}^{3}} & {{b}^{2}}+ab+{{a}^{2}} & {{c}^{2}}+ac+{{a}^{2}} \\ \end{matrix} \right| $
$=(b-a)(c-a) \times\left[\left(c^{2}+a c+a^{2}\right)-\left(b^{2}+a b+a^{2}\right)\right]$
$=(b-a)(c-a)\left[c^{2}-b^{2}+a c-a b\right]$
$=(b-a)(c-a)(c-b)[c+b+a]$
Note: The value of the determinant does not change either by any rows or by any columns operation.